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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink]
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Re: Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink]
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motion2020 wrote:
Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}
Set B includes numbers the same as in the set A with 3 subtracted from each number and then multiplied by 1.5



Quantity A
Quantity B
7
Arithmetic average of the set B


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Average of set A \(= \frac{4+5+5+5+5+5+5+5+6+7+11+12+16}{13} = 7\)

If we subtract 3 from each value, then the new average (arithmetic mean) = 7 - 3 = 4

If we now multiply each value by 1,5, the new average = (4)(1.5) = 6

So, the average of Set B = 6

So we get:
QUANTITY A: 7
QUANTITY B: 6

Answer: A
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink]
and morale of this question is that adding and multiplying or dividing the mean (average) by constant should change the mean (new mean is obtained) by that constant's addition, multiplication or division
GreenlightTestPrep wrote:
motion2020 wrote:
Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}
Set B includes numbers the same as in the set A with 3 subtracted from each number and then multiplied by 1.5



Quantity A
Quantity B
7
Arithmetic average of the set B


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Average of set A \(= \frac{4+5+5+5+5+5+5+5+6+7+11+12+16}{13} = 7\)

If we subtract 3 from each value, then the new average (arithmetic mean) = 7 - 3 = 4

If we now multiply each value by 1,5, the new average = (4)(1.5) = 6

So, the average of Set B = 6

So we get:
QUANTITY A: 7
QUANTITY B: 6

Answer: A
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink]
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